When we move on to n=3 variables, we now have, as basic building blocks,

These are just the coefficients of in the expansion of . Once again, any symmetric polynomial in x, y, z with integer coefficients can be expressed as a polynomial in P, Q and R with integer coefficients.

Example. If we let , then using the technique described in the previous post, we obtain the recurrence relation:

Starting from , , , we obtain:

;

…

Let’s solve some problems now.

Example 1. Factor the polynomial .

Solution. Since that’s a symmetric polynomial, we can express it as a polynomial in P, Q and R. From the above, we have , so:

.

So . ♦

Note. To be fair, this method worked only because the individual factors were also symmetric. Sometimes one gets something like the following.

Example 2. Let and . Express the polynomial as a polynomial in P, Q, R.

Solution. No, we’re not going to expand the monster. Instead, we note that it is homogeneous of degree 6. Since P, Q and R are homogeneous of degrees 1, 2, and 3 respectively, we must have:

for some constants . That’s too much to solve linearly, so we prune down the cases further by singling out one variable: x. Indeed, the degree of x in A is 4, with leading coefficient . So:

and can’t appear since they contain higher powers of x, i.e. ;

also, only appears in and , with respective leading terms and . So .

with only three unknowns left, we can substitute explicit values into x, y, z to solve for them, or we can substitute x=y to first weed out from the coefficient of .

It turns out so we get

Exercise 1. Factor the polynomial A. [Hint (highlight to read): deduce from the above that the factors aren’t symmetric, so given any factor, we can obtain new ones by permuting the variables. Such a factor has degree at most 2, let z=0 and conclude the degree is in fact exactly 2. Guess the factor and prove it works. ]

Example 3. Solve the system of simultaneous equations:

where x, y and z are distinct complex numbers.

Solution. Let s = x+y+z and rewrite the above equations as:

Hence, taking s as a constant, we see that x, y and z are all roots of the cubic equation , where T is unknown. Since x, y and z are distinct, these are precisely all the roots. Expanding the polynomial gives and the sum of the roots is thus given by 3s. Yet the sum of the roots is also x+y+z = s. Thus s=0 and x, y and z are roots of . So and other permutations. ♦